java - 如何设置 jackson 为假时忽略 boolean 属性？

Question

我使用 Jackson，ObjectMapper.readValue(json, Class)。

上课:

Component {
  private String name;
  private String someField;
  private boolean show = true; // if false -> skip it object
}

和可扩展类:

ExtendedComponent extends Component {
  private List<Component> components = emptyList();// if all of object in list not showed -> skip field
}

和复杂类:

ComplexComponent extends Component {
      ExtendedComponent component;
    }

和 json:

{
  "complexComponent": {
    "name": "complexName1",
    "show": true,
    "someField": "complex",
    "extendedComponent": {
      "components": [
        {
          "name": "someName1",
          "show": true,
          "someField": "someField"
        },
        {
          "name": "someName2",
          "show": false,
          "someField": "someField"
        },
        {
          "name": "someName3",
          "show": true,
          "someField": "someField"
        }
      ]
    }
  }
}

ExtendedComponent 中如何只获取名称为 someName1 和 someName3 的对象？

如果未显示所有组件，如何获取可为空的 ComplexComponent？

Answer 1

取决于你的 jackson 2 版本

@JsonInclude(Include.NON_DEFAULT)

或

@JsonInclude(Include.NON_EMPTY)

应该可以解决问题。